THe U238 nucleus is unstable and does undergo "ordinary" radioactive decay.
For a U238 nucleus to undergo fission you need to provide the nucleus with something akin to an activation energy of about 7 MeV.
One model is that the nucleus is like a liquid drop and adding a neutron can make the drop oscillate so much that it breaks up into two smaller pieces but to initiate a large oscillation requires an input of energy.
An extra neutron introduced into the nucleus provides about 6 MeV of this required amount of energy from the extra binding energy between the nucleus and the neutron. If an extra 1 MeV or more of energy is provided by kinetic energy of the neutron then the nucleus will be able to undergo nuclear fission.
One of the reasons that U235 is favoured in fission reactors is that it contains an odd number of neutrons and the pairing of neutrons within the nucleus is energetically favourable. So an incoming neutron does not have to provide any kinetic energy because the 1 MeV is provided by the neutron pairing process.
If you look at the periodic table you will note that as the nuclei get bigger the neutron to proton ratio increases. So larger nuclei are richer in neutrons than smaller nuclei.
When fission occurs the neutron rich uranium produces small nuclei which have many too many neutrons and so are very unstable.
So some of these excess neutrons are flung out during the fission process (and then can be used to initiate further fissions) and the rest are converted within the nuclei of the fission fragments into protons by beta minus decay until stable nuclei are formed.